It is obvious that [itex]C\subseteq ccone(K+C)[/itex]. Assume that this inclusion was strict, then there would be a direction d which is not in C. This d has a >0 distance from C. Thus the multiples of d grow further away from C. That is, the distance from d to C becomes arbitrarily big. But we still have that d is in ccone(K+C). Can you find a contradiction with that?

Thanks for your reply :) and yes, a characteristic cone is the same as a recession cone.
Then I must show that T = S

I can show that if a belongs to S, then a must belong to T as well.
Let a=y+d belong to C.
if y belongs to C, and because C [itex]\subseteq[/itex] C+K, then y must belong to C+K
Therefore a=y+d must belong to C+K.

I think you made a mistake in your picture since T and S are exactly the same there.

But I see what you mean. Let's prove this in steps. Let's begin with this: let d be a direction not in C. Can you prove that the distance between x+rd and C becomes arbitrarily large as r becomes large?

I.e. can you show that [itex]d(x+rd,C)\rightarrow +\infty[/itex] as [itex]r\rightarrow +\infty[/itex]?

Do you see intuitively why it must be true?
Consider for example the cone [itex]C=\{(x,0)~\vert~x\in \mathbb{R}\}[/itex] in [itex]\mathbb{R}^2[/itex]. Take something not in C, for example (1,1). Do you see that multiples of (1,1) are getting further away from C? That is, if [itex]r\rightarrow +\infty[/itex], then the distance between (r,r) becomes arbitrarily large.

Well, x+rd is getting further away from C. But if d is in ccone(K+C), it must hold that x+rd is in K+C. And thus we must be able to express x+rd=k+c. But as the distance between x+rd and c becomes large, then k must become large. Thus K must be unbounded.

Take d in C, then for all x in C, we have that x+rd is in C. In particular rd is in C.
Now, take c+k in C+K, then c+k+rd=k+(c+rd) is in C+K (since C is convex). Thus for every x in C+K, we have that c+rd is in C+K